fender deluxe 112 plus -2nd problem - drive channel

[cwpeters]

Thanks for all the help on my last post.  I fixed the problem with the hum.  And I was wrong and felt like a dumbass.  I did put the caps in backwards.  Ive been trying to teach myself electronics for about a 1 1/2 years and even though I thought I understoud electrolytic caps, I was thinking the - stripe was the + indicator.  Lesson learned, I'll never make that mistake again.   Luckily, I didnt wreck the caps or anything else, I hope.  I put them in the "CORRECT"  way and the hum is gone.  It sounds clean and crisp.  However, The drive channel switch doesnt change from drive to normal. It is stuck on drive. I took the switch out, checked it and it "seems" fine.  Is there anything else that it could be?  I'm thinking that it has to be the switch.  When I push it, nothing happens.  Any advice.

[teemuk]

You need a schematic. The switch is just controlling the input of a logic circuit – which in turn controls the state of several JFET switches. You have something in that circuit broken down in a way that forces the switching logic to think the "drive" state is on.

[cwpeters]

here is the schematic.  By looking at the schematic do you see anything I should check? And how should I test it?  http://www.bnv-gz.de/~ooehmann/dexer.php?d=fender

[teemuk]

You have the test point TP20 from which you can verify if the logic is working when you operate the switch either by using footswitch or the amp's internal switch. The test point likely stays at steady +15V no matter what. The LEDs will likely indicate the same thing.

U4A is a comparator: Its non-inverting input should sit at about 0.6 V (diode junction of CR20 or CR26?) and whenever the input signal at inverting input is lower than this the output of the comparator is about 15 V. If inverting input has higher voltage potential than the non-inverting input, the comparator changes its state and assumes a negative (-15V) output.

The comparator controls the JFETs that do the actual switching. Check if it is working correctly: For example, If diode at non-inverting input is open or there is no ground reference (cold solder) this would force the output to steady 15V. You could as well just have a broken comparator opamp.

How does the switch work? ....As you can see, the footswitch (and the internal drive switch) are fed with 31.4V AC signal from the power supply. This allows switching both reverb and drive while using a footswitch wire with only 2 conductors. Shunting the AC signal through a diode (or set of them), which effectively limits the amplitude, controls the input of the comparator. There is an individual shunt control for each half of the wave. Each half of the wave controls its equivalent function: The positive half wave controls the drive, the negative half wave the reverb. The drive is on when positive half wave is limited. This is done either by "activating" shunt diode CR17 – or the shunt LED and diode inside the footswitch. The main point is that the switching limits the AC amplitude from 31.4V to about few volts or less. Check if this happens, check if you even have AC signal there.

R83 and R84 attenuate the rather high AC voltage to make it more suitable for the comparator. The attenuation is about 10.

CR18, C45 and R85 form a (half wave) rectifier for the positive half wave: In order to have the switching logic work correctly the AC signal from the switching units must be rectified to steady DC. There is a similar half wave rectifier for the negative half, which controls the reverb switch. We can ignore this circuit but it basically works similarly. Without footswitch in place the reverb is constantly "switched" by shunt CR15.

Note that the rectifier introduces another diode forward voltage drop. Now, the circuit is either rectifying about 3V AC or something less. I derived that figure from 31.4V/10. This is of course the amplitude of the positive half wave only and we are ignoring the existence of the negative half since it makes no difference here. If the rectified signal is 3V we can deduce that 3V-0.6V is still way more positive than 0.6V at the non-inverting input of the comparator so the comparator assumes a -15V output and the drive is off. If the drive switch shunts the diode, the input amplitude is very low, this leads to maybe few millivolt input signal at inverting input (which is less than 0.6V > comparator assumes output of +15V > "drive" is on). Something in this chain could be wrong as well. Again, testing the comparator's inputs should tell you this.

Like I said, very likely the comparator is forced to a certain state. It's up to you to measure if it is and then find out why. I guess you understand that there are few possible fault scenarios. IMO, it's not worthwhile to guess them here as you can get better results by simply measuring and seeing if something is wrong by yourself.

If the logic works correctly then the JFETs are the culprit. However, I'm pretty sure you will find something fishy from the logic circuit.

I hope this helped...